# Lecture 8: Projective Isometries

We have a properly convex domain with a Hilbert metric. We define the distance between two points \(x,y\) by \(d_{\Omega}(x,y) = log([a:x:y:b])\). If we have a projective transformation that preseves cross ratios we call it an isometry of \(d_{\Omega}\).

If we restrict ourselves to matrices with determinant plus or minus one \(SL^{\pm} (\Omega)\) and then lift to the sphere (since the double cover of a projective space \(RP^{n}\) is a sphere \(S^n\)) and choose one of the preimages \(\tilde{\Omega}\) with the property that \(\tilde{\Omega}\) maps to itself it becomes much simpler to talk about eigen values and classify isometries.

\(A\) is called elliptic if there exists \(p \in \Omega\) such that \(Ap=p\). Here \(p=[v]\) the equivalence class of some vector i.e. \([Av]=[v]\). This implies that \(Av = \lambda v\) for some \(\lambda > 0\). Hence fixed points coresspond to eigenvectors of the matrix with positve eigenvalues. \(A\) is parabolic if every eigenvalue of \(A\) has modulus \(1\), but \(A\) is not elliptic. Finally we say that \(A\) is parabolic otherwise.

We define the translation length \(t(A)=\inf_{x \in \Omega} d_{\Omega}(x,Ax) \geq 0\) and also define the \(minset(A)=\{x \in \Omega \mid d_{\Omega}(x,Ax) = t(A)\}\) of an isometry. In Hyperbolic geometery \(H^{n}\) \(t(A) > 0\) if and only if \(A\) is hyperbolic. In this case points along the geodesic are moved the minimum amount i.e. the minset lies in some geodesic, called the axis. If \(t(A)=0\) then \(A\) is elliptic if and only if the minset is nonempty and parabolic (along with a unique point \(p \in \partial \tilde{\Omega}\) called the parabolic fixed point) if and only if the minset is empty.

Proposition: Let \(A \in SL(\Omega)\), then \(t(A)=log\vert\lambda / \mu\vert\) where \(\lambda\) is an eigenvalue of largest modulus and \(\mu\) is an eigenvalue of smallest modulus.

Proposition: \(A \in SL(\Omega) \subset SL^{\pm}(n+1,R)\) is elliptic if and only if \(A\) is conjugate onto \(O(n+1)\)

Proof: By the definition of elliptic there is some point in the domain \(p \in A\) that is fixed by the elliptic. Here we have a diffeomorphism so \(dA\) acts on \(T_p \simeq R^{n}\). \(d_{\Omega}\) is a norm on \(T_p(\Omega)\) specifically a Finsler metric. \(dA\) is an isometry of \(T_p(\Omega)\), which implies there exists an inner product \(\beta\) on \(T_p(\Omega)\) such that \(dA\) is an isometry of \(\beta\). Implies \(dA \in Isom(\beta)=O(n)\). (We did not finish proof in lecture, but this implies the conclusion apparently.)

\(\bar{\Omega}\) is homeomorphic to some \([0,1]^n\) so any isometry of \(\Omega\), \(A \in SL(\Omega)\) it preserves the closure \(A(\bar{\Omega})=\bar{\Omega}\). The Brower fixed point theorem implies that there exists \(p \in \bar{\Omega}\) such that \(Ap=p\).

Case1: The fixed point is in the interior \(p \in \Omega\) then \(A\) is an elliptic.

Case2: The fixed point is on the boundary \(p \in \partial \bar{\Omega}\). (In addition it fixes a hyperplane containing that point on the boundary.)

Define a supporting hyper plane \(H\) to \(\Omega\) at \(p \in \partial \bar{\Omega}\) is a projective hyperplane \(H=P(V)\) with \(dim(V)=n\), such that \(p \in H\) and \(H \cap \Omega = \emptyset\).

Exercise: Supporting hyperplanes exist.

Proposition: The set of supporting hyperplanes \(\{P(ker \phi) : [\phi] \in CC(RP^n)^*\}\) is dual to a nonempty compact convex set in \((RP^n)^*\).

Proof: \(\Omega \subset RP^n\) and \(\Omega^* \subset RP^n\) a linear map is in the dual domain if and only if \(P(ker \phi)\) misses the interior of the domain, \(\Omega^* = P(\{\phi \in V^* \mid \phi(\Omega) >0\})\). \(\Omega\) must be in \(S^n \subset R^{n+1} = V\) for this to make sense, so we have to lift to the double cover (\(V\) a vector space).

Corollary: If \(A \in SL(\Omega)\) fixes a point in the boundary of the ball \(p \in \partial \Omega\) then there exists a supporting hyperplane \(H\) to \(\Omega\) at \(p\) such that \(A(H)=H\)

Proof: \(H\) corresponds to a point in a dual compact convex set \(C\). \(A\) sends supporting hyperplanes to supporting hyperplanes, so \(A: C \rightarrow C\). Then by the Brower Fixed Point Theorem there exists \([\phi] \in C\) such that \(A[\phi]=[\phi]\). Therefore \(H=ker \phi\) is a supporting hyperplanefixed by \(A\).

Structure of fixed set of \(A \in SL(\Omega)\): Let \(p=[x]\) for \(x \in R^{n+1}\), \(A\) fixes \(p\) if and only if there is a positive number \(\lambda\) such that \(Ax=\lambda x\). In other word equivalent to \(x\) being and eigenvector with a positive eigenvalue. Given an eigenvalue \(\lambda>0\) for \(A\) we have the eigenspace \(E_\lambda\). Define the Fixset \(Fix(\lambda, A, \bar{\Omega})= \bar{\Omega} \cap P(E_{\lambda})\) this will be compact and convex as both \(\bar{\Omega}\) and \(P(E_{\lambda})\) are. The set of all fixed points will be \(Fix(A,\Omega) = \bigcup_{\lambda > 0} Fix(\lambda, A, \Omega)\).

Example: Suppose \(\Omega = \Delta\) then

\[SL(\Omega) = \{ ()~a,b > 0 \} \rtimes \{permutation~matrices\}\]Consider \(A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 0 \\ 0&0&1/4 \end{pmatrix}\) then the fixed points we have are \(Fix(\lambda =2,A,\Delta)\) and \(Fix(\lambda = 1/4,A, \Delta)\).

Dynamics of \([A] \in P_{+}GL(n+1,R)\):

First of all the largest eigenvalues will ‘win’ (i.e. they will blow up the fastest as a matrix is multiplied by itself repeatedly) hence we need only consider the largest eigenvalues. Consider \(A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 2 \\ 0&0&2 \end{pmatrix}\) then \(A^n=2^n \{ \begin{pmatrix} 1&0&0 \\ 0 & 1 & n \\ 0&0&1 \end{pmatrix}\). From this we can see that the largest Jordan block will blow up fastest therefore we will just talk about the dynamics of a single \(k+1 \times k+1\) Jordan block. The growth of some block is \(A^{n}=\lambda^n (I + nN + \frac{n(n-1)}{2}N^2+ ... + (\frac{n}{k}N^k))\) the last term wins as \(N \rightarrow \infty\). We define the power of a Jordan blonck with eigenvalue \(\lambda \in C\) is \((\vert\lambda\vert,k+1)\) and we use the lexicographical order. We get a partial ordering on Jordan blocks from this.

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